David Bernier wrote:>>Let's suppose the base field is Q, and P(x) is an irreducible>polynomial of degree n over Q. Let alpha_1, ... alpha_n>be the n conjugate roots in the splitting field L (subfield of>C, the complex numbers) of P(x) over Q.>>If sigma: {alpha_1, ... alpha_n} -> {alpha_1, .. alpha_n}>is a permutation of the n conjugate roots,>>then according to me if a field automorphism of phi of L exists>which acts on {alpha_1, ... alpha_n} the same way the >permutation sigma does,all the elementary symmetric polynomials >in n indeterminates must be invariant under the application of >such elementary symmetric polynomials:>>[wikipedia, with def. of elementary symmetric polynomials]>>http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial>>In the other direction, if we have a sigma, permutation as above,>and all the elementary symmetric polynomials are left>invariant, does it follow that for the splitting field L,>there is a field automorphism phi of L such that> phi(alpha_j) = sigma(alpha_j), 1<=j<=n ?>In other words, phi acts on the alpha_j the same way sigma >does.>>If the elementary symmetric polynomials are left invariant >by sigma, does it follow that some automorphism phi of L>acts on {alpha_1, ... alpha_n} the same way sigma acts ?

The elementary symmetric functions of the roots are left invariant by _any_ permutation of the roots.